Question

我有一个由3个子图组成的图。我想在第二行的中间找到最后一个子图。当前，它位于图的左下方。我该怎么做呢？我找不到堆栈溢出的答案。

    fig, axes = plt.subplots(nrows=2, ncols=2, figsize=(13,10))
    ax= axes.flatten()
    ax[0].plot(vDT, np.cumsum(mWopt0[asset0,:])*percentage/iTT, label= 'COAL, c = 0')
    ax[0].legend()
    ax[0].set_title('Proportion in most invested stock')
    ax[1].plot(vDT, np.cumsum(mWopt01[asset01,:])*percentage/iTT, label= 'OINL, c = 0.1')
    ax[1].plot(vDT, np.cumsum(mWopt03[asset03,:])*percentage/iTT, label= 'OINL, c = 0.3')
    ax[1].plot(vDT, np.cumsum(mWopt05[asset05,:])*percentage/iTT, label= 'OINL, c = 0.5')
    ax[1].plot(vDT, np.cumsum(mWopt2[asset2,:])*percentage/iTT, label= 'OINL, c = 2')
    ax[1].plot(vDT, np.cumsum(mWopt5[asset5,:])*percentage/iTT, label= 'OINL, c = 5')
    ax[1].plot(vDT, np.cumsum(mWopt10[asset10,:])*percentage/iTT, label= 'OINL, c = 10')
    ax[1].legend()
    ax[1].set_title('Proportion in most invested stock')
    ax[2].plot(vDT, np.cumsum(mWopt01[index,:])*percentage/iTT, label= 'c = 0')
    ax[2].plot(vDT, np.cumsum(mWopt01[index,:])*percentage/iTT, label= 'c = 0.1')
    ax[2].plot(vDT, np.cumsum(mWopt03[ index,:])*percentage/iTT, label= 'c = 0.3')
    ax[2].plot(vDT, np.cumsum(mWopt05[index,:])*percentage/iTT, label= 'c = 0.5')
    ax[2].plot(vDT, np.cumsum(mWopt2[index,:])*percentage/iTT, label= 'c = 2')
    ax[2].plot(vDT, np.cumsum(mWopt5[index,:])*percentage/iTT, label= 'c = 5')
    ax[2].plot(vDT, np.cumsum(mWopt10[index,:])*percentage/iTT, label= 'c = 10')
    ax[2].legend()
    ax[2].set_title('Proportion invested in index')
    ax[0].set_ylabel('Expanding window weight')
    ax[1].set_ylabel('Expanding window weight')
    ax[2].set_ylabel('Expanding window weight')
    ax[3].remove()
    fig.autofmt_xdate(bottom=0.2, rotation=75, ha='right')
    plt.savefig('NSE_por_unrestricted_mostweightSI.jpg', bbox_inches='tight')
    plt.show()

Answer 1

matplotlib.gridspec.Gridspec解决了您的问题，可以传递给plt.subplot。在this answer中，您可以看到可以使用4x4网格轻松将图放置在中间：

import matplotlib.gridspec as gridspec
import matplotlib.pyplot as plt

gs = gridspec.GridSpec(4, 4)

ax1 = plt.subplot(gs[:2, :2])
ax1.plot(range(0,10), range(0,10))

ax2 = plt.subplot(gs[:2, 2:])
ax2.plot(range(0,10), range(0,10))

ax3 = plt.subplot(gs[2:4, 1:3])
ax3.plot(range(0,10), range(0,10))

plt.show()

您可以在此处查看有关gridspec的演示：https://matplotlib.org/tutorials/intermediate/gridspec.html#sphx-glr-tutorials-intermediate-gridspec-py

唯一的问题是您使用的是fig, axes =模式，我不认为它通常与Gridspec一起使用。您需要重构一下。

将第三个子图居中于第二行python的中间

1 个答案: